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\begin{document}
{\today}
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%\input{Abstract}
%\frontmatter
%\input{Preface}
%\mainmatter
\tableofcontents
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%\input{Varieties}
%\input{Schemes}
%\color{black}
%\input{Sheaves_on_Grothendieck_Topology}
%\input{Category_of_Abelian_Sheaves}
\section{Flatness and Ramifications}
\begin{definition}[Family of schemes]
We define a family of schemes parametrised by the scheme $B$, or with base B, to be a morphism of schemes $\pi:X\rightarrow B$
\end{definition}
The family of schemes in the previous definition is thought of as the family of the fibers $\pi^{-1}(b),\forall b\in B$.
%%%%\begin{remark}The above definition is too general, so the family does not have \tcb{anything in common} For any family of schemes $\pi:X\rightarrow B$, and $b$ closed in $B$. One might consider the family $\pi\prime:X\backslash\pi^{-1}(b)\bigsqcup Y\rightarrow B$, for some scheme $Y$, defined by $\pi\prime|_{X\backslash\pi^{-1}(b)}=\pi|_{X\backslash\pi^{-1}(b)}$, and sending $Y$ to $\{b\}$.\end{remark}
The aim of this section is to make sense of continuity of a family of schemes.
\subsection{Limits}
\begin{definition}[A Family of closed subschemes of a given scheme]
We define a family of closed subschemes of a scheme $A$ over a base $B$ to be the family $\pi:X\rightarrow B$, where $X\subseteq B\times A$ closed subscheme of $B\times A$, and $\pi=\pi_B|_X$, the restriction of the projection $\pi_B:B\times A \rightarrow B$. I.e. $\pi:X\hookrightarrow A\times B\stackrel{\pi_B}{\rightarrow} B$.
\end{definition}
\tcb{Then, the fibers $\pi^{-1}(b)$ are closed subschemes of the fibers $A_b=\pi_B^{-1}(b)$}
\begin{definition}[Limit of a Family of closed subschemes]
Let $B$ be a \tcb{non-singular, one-dimensional scheme}, $0\in B$ a closed point in $B$, $X^{\ast}$ a family of closed subschemes of a scheme $A$ over a base $B^{\ast}=B\backslash \{0\}$. Then, we define the limit of the $\displaystyle\lim_{b\rightarrow 0}X^{\ast}_b$ to be the fiber of $0$ of the family $\overline{X^{\ast}}\stackrel{i}{\hookrightarrow} A\times B \rightarrow B$
\end{definition}
%%%%\tcb{so far limit are taken for subschemes of the affine space}\\
%%%\tcb{prove that the extension of the above morphism of schemes is a morphism of schemes}.\\
%%%%If $B=\Spec R$ affine scheme, $\gom=<t>$ corresponding to $0\in B$. Then, $B^{\ast}=\Spec R[t^{-1}]$ $X^{\ast}\subseteq\AF^n_{B^{\ast}}$ with $I(X^{\ast})\subseteq \Spec R[t^{-1}][x_1,...,x_n]$, then the ideal of the subscheme $X\subseteq \AF^n_B$ is $I(X)=I(X^{\ast})\bigcap R[x_1,...,x_n]$\\
%%%Straightforward definition of limit.
%%%%examples of limits
%%%contnuous family
%%%%Serre introduced the notion of flatness to extend the notion of continuous family of subschemes into more general settings.
\begin{definition}
A module $M$ over $R$ is flat if the functor $M\otimes_R - $ maps monomorphisms to monomorphisms.
\end{definition}
\begin{definition}
The family $\pi:X\rightarrow B$ of schemes is flat if $\forall x\in X$ the local ring $\bcO_{X,x}$ is flat as $\bcO_{B,\pi(x)}$-module, via $\pi^{\#}$.
\end{definition}
%%%%What does the flat family of schemes have in common.??
\begin{example}[Eisenbud Ex P 72-73]
Let $K$ be an algebraically closed field, $R=K[t]\otimes_K K[x,y,z]$, $Q_1=z(z-tx), Q_2=z(z-ty), Q_3=(z-tx)(z-ty)\in R$ $\goa=(Q_1,Q_2,Q_3)R\subseteq R$, $X=\Spec{(R/\goa)}$, $A=\Spec(K[x,y,z])$ and $B=\Spec(K[t])$. Consider the family of schemes $\pi:X\stackrel{i}{\hookrightarrow} B\times A \stackrel{s}{\twoheadrightarrow} B$ induced by the $K[t]\stackrel{i}{\hookrightarrow} R \stackrel{s}{\twoheadrightarrow} R/\goa$.\\
Let $\underline{0}=(t)\in B$, then $B^{\ast}=B\setminus\{\underline{0}\}=\Spec K[t]_t$. Then, considering the coproduct diagram:
$$
\xymatrix{K[t]\ar@{^(->}[r]^i\ar@{^(->}[d]_i& R\ar@{^(->}[d]_i \ar@{->>}[r] & R/\goa\ar@{^(->}[d]_i\\
K[t]\ar@{^(->}[r]^i& R_t \ar@{->>}[r] & (R/\goa)_t
}
$$ 
Where $(R/\goa)_t\cong R_t/\gob$, where $\gob$ is generated by the embeddings of $Q_1,Q_2,Q_3$ in $R_t$. Then we have the family, $$\pi^{\ast}:X^{\ast}=\Spec R_t/\gob\stackrel{i}{\hookrightarrow} B^{\ast}\times A \stackrel{s}{\twoheadrightarrow} B$$
Therefore, in order to find the limit $\displaystyle\lim_{b\rightarrow \underline{0}}$, we need to find the closure of the embedding of $X^{\ast}$ in $A\times B\cong \AF^4$. Then, considering the canonical morphisms $s\circ i:R\rightarrow R_t\rightarrow R_t/\gob$, we find that $\overline{i(X^{\ast})}=\Spec (R/\ker(s\circ i))=\Spec(R/i^{-1}(\gob)=\Spec(R/(\gob\cap R)$, \tcr{$\gob=\displaystyle\bigcap_{\gop\supseteq \goa}\gop$}.

Compute $\gob\cap R$

We are abusing the notation, by $\gob\cap R$, we mean the preimage of $\gob$ of the embedding $R\hookrightarrow R_t$, also we are using the same notation for element of $R$ and there image of the previous embedding.\\
We defined $\gob=(Q_1,Q_2,Q_3)R_t$, then $Q_1,Q_2,Q_3 \in \gob$, since $Q_1,Q_2,Q_3\in R$. Hence, $Q_1,Q_2,Q_3\in \gob \cap R$.\\
$\frac{1}{t}\in R_t$, hence $Q_4:=\frac{Q_1-Q_3}{t}=y(z-tx),Q_5:=\frac{Q_2-Q_3}{t}=x(z-ty),Q_6:=\frac{xQ_4-yQ_5}{t}=xy(y-x)\in \gob$. We notice that $Q_4,Q_5,Q_6\in R$. Hence, $Q_4,Q_5,Q_6\in \gob\cap R$. Therefore, $I:=(Q_1,Q_2,Q_3,Q_4,Q_5,Q_6)R\subseteq\gob\cap R$. \tcr{The task is that we were working on is to show the opposite inclusion}.\\
Let $h\in \gob\cap R$, then $h=\frac{c_1Q_1+c_2Q_2+c_3Q_3}{t^n}$, for some $c_1,c_2,c_3\in R, n\in \NN$. then we need to show that $h\in I$
\tcr{we were trying to prove it by induction on $n$}.\\
\tcb{before we start we notice the following relations:}\\
$yQ_1=zQ_4,xQ_2=zQ_5,zQ_3=(z-tx)Q_2=(z-ty)Q_1,Q_1=tQ_4+Q_3,Q_2=tQ_5+Q_3, zQ_6=xy(Q_5-Q_4)$\\
For $n=0$, there is nothing to prove.\\
For $n=1$, $th=c_1Q_1+c_2Q_2+c_3Q_3=c_1(tQ_4+Q_3)+c_2(tQ_5+Q_3)+c_3Q_3=t(Q_4+Q_5)+(c_1+c_2+c_3)Q_3\in R$. $(c_1+c_2+c_3)Q_3$ is divisible by $t$. $t$ is irreducible in $R$, and $t\nmid Q_3$. Then, $c_1+c_2+c_3=t c$, for some $c\in R$.\\
Hence, $th=t(cQ_3+c_1Q_4+C_2Q_5)$, then $h=cQ_3+c_1Q_4+C_2Q_5\in I$.\\
For $n=2$, We find $th=c_1Q_4+c_2Q_5+cQ_3, c_1+c_2+c_3=tc$. Then to find further conditions on $c_1,c_2,c_3$ we may substitute $Q_4,Q_5,Q_3$ to find that: $th=c_1(y(z-tx))+c_2(x(z-ty))+c((z-tx)(z-ty))=t(ctxy-c_1yx-c_2xy-czx-czy)+z(c_1y+c_2x+cz)$.\\
Hence, $c_1y+c_2x+cz=td$, for some $d\in R$ \tcb{did not manage to use this condition  to show that $h\in I$!?}\\

We can assume that $c_3\in K[x,y,t]$, i.e. does not have terms with $z$, that $zQ_3\in (Q_1)\cap(Q_2)$.\\

We can show that $zh\in I$, that $zth=c_1zQ_4+c_2zQ_5+czQ_3=c_1yQ_1+c_2xQ_2+(td-c_1y-c_2x)Q_3=c_1y(Q_1-Q_3)+c_2x(Q_2-Q_3)+tdQ_3=c_1ytQ_4+c_2xtQ_5+tdQ_3$. Hence, $zh==c_1yQ_4+c_2xQ_5+dQ_3\in I$, substituting $Q_4,Q_5,Q_3$, we find, after dividing by $z$, that $h=-c Q_5-cQ_4-ctxy+dz-xy(c_1+c_2)$ \tcb{we should find further conditions on $c_1,c_2,c_3$ to be able to show that $h\in I$.}


Another approach


Let $h\in I$, i.e. $h=a_1Q_1+a_2Q_2+a_3Q_3+a_4Q_4+a_5Q_5+a_6Q_6$, then multiplying by $t^2$, and substituting $Q_4,Q_5,Q_6$, we find $t^2h=c_1Q_1+c_2Q_2+c_3Q_3$, where:\\
$c_1=t^2a_1+ta_4+xa_6$\\
$c_2=t^2a_2+ta_5-ya_6$\\
$c_3=t^2a_3-ta_4-ta_5-xa_6+ya_6$\\
Then in order to have $I=\gob\cap R$, we need to find \tcb{canonical conditions} on $c_1,c_2,c_3$ such that $a_1,...,a_6$ could be \tcb{always} retrieved from $c_1,c_2,c_3$. If such conditions found, then $a_6$ is terms of $c_1$ that is not divisible by $t$, divided by $x$.....\\
Based on the above relations on $Q$'s, we may assume that $y\nmid a_1,x\nmid a_2,x\nmid a_3,y\nmid a_3,z\nmid a_3,z\nmid a_4,z\nmid a_5,z\nmid a_6$. Also, we may assume $z\nmid c_3$, moreover we may require the coefficients not to include any term of these variables that does not divide it...
\tcb{these conditions is not enough, if order to have $I=\gob\cap R$, we need to have $t\mid yc_1+xc_2$..}

\tcb{the rest is either wrong calculations, or examples of such $h$ in order to find relation between $a$'s and $c$'s...}
\end{example}


%\input{CounterExample}
%\backmatter
\appendix
%\input{Categories}
\section[Commutative Algebra]{Commutative Algebra}
In this section, let $R$ a commutative unitary ring, unless mentioned otherwise.
\subsection{Projective Modules}
\subsection{Tensor Product}
\subsection{Flat Modules}[Tensor Exact]
\begin{lemma}
Let $M\in R-\Mod$, then the following conditions are equivalents:
\begin{itemize}
\item For every exact sequence $\xymatrix{E'\ar[r]^f & E \ar[r]^g & E''}$ in $R-\Mod$, the sequence $\xymatrix{M\otimes_R E'\ar[r]^{M\otimes_R f} & M\otimes_R E \ar[r]^{M\otimes_R g}& M\otimes_R E''}$ is also exact.
\item For every short exact sequence $\xymatrix{0\ar[r]&E'\ar[r]^f & E \ar[r]^g & E''\ar[r]&0}$ in $R-\Mod$, the sequence $\xymatrix{0\ar[r]&M\otimes_R E'\ar[r]^{M\otimes_R f} & M\otimes_R E \ar[r]^{M\otimes_R g} & M\otimes_R E''\ar[r]&0}$ is also exact.
\item For every exact sequence $\xymatrix{0\ar[r]&E'\ar[r]^f & E}$, i.e. a monomorphism, the sequence $\xymatrix{0\ar[r]& M\otimes_R E'\ar[r]^{M\otimes_R f} & M\otimes_R E}$ is also exact, i.e. a monomorphism.
\end{itemize}
\end{lemma}
\begin{proof}
\tcg{type it}
\end{proof}
\begin{definition}
Let $M\in R-\Mod$,then we call $M$ a flat module over $R$, or tensor exact module, if any of the conditions of the previous lemma holds.
\end{definition}
\begin{example}[Examples of flat modules].
\begin{itemize}
\item $R$ is flat over itself.
\item Let $M=\oplus_{i\in I}M_i$ in $R-\Mod$. Then, $M$ is flat iff $M_i$ is flat $\forall i\in I$. Hence, free modules, over $R$, are flat over $R$.
\item Projective modules in $R-\Mod$, are flat over $R$.
\end{itemize}
\end{example}


\begin{proof}
\tcg{type it}
\end{proof}
\begin{proposition}
\begin{itemize}
\item Let $S$ be a multiplicative subset of $R$. Then $S^{-1}R$ is flat over $R$.
\item A module $M$ is flat over $R$ iff $M_{\gop}$ is flat over $R_{\gop}$ for every prime ideal $\gop$ of $R$.
\item Let $R$ be a principle ideal domain. Then, $M$ in $R-\Mod$ is flat over $R$ iff it is torsion free.
\end{itemize} 
\end{proposition}
\begin{proof}
\tcg{type it}
\end{proof}


\begin{counterexample}
Let $R$ be a n integral domain, if $M \in R-\Mod$ has a torsion, over $R$, then $M$ is not flat, over $R$.
\end{counterexample}
\begin{proof}
Let $M \in R-\Mod$ has a torsion $m_0\in M$, over $R$. Let $r\in R$ be the regular element annihilating $m$. Then we have the embedding monomorphism $\xymatrix{0\ar[r]& R\ar[r]^i & R_r}$, where $R$ and $R_r$ are considered in $R-\Mod$. Then we have the sequence $\xymatrix{0\ar[r]& M\otimes_R R\ar[r]^{M\otimes_R i} & M\otimes_R  R_r}$. Notice that $m\otimes 1 \neq 0 \in M \otimes_R R\cong M$, whereas ${M\otimes_R i}(m\otimes 1)=m\otimes \frac{1}{1}=r\cdot m\otimes \frac{1}{r}=0\otimes \frac{1}{r}=0 \in M\otimes_R  R_r$. Hence, ${M\otimes_R i}$ is not injective.\\
For example, consider $R=\ZZ$, and $M=\ZZ\diagup{2\ZZ}$.
\end{proof}

\begin{lemma}[Snake lemma]

Let $F$ be a flat module, over $R$, and let $\xymatrix{0\ar[r]&N\ar[r]^f & M \ar[r]^g & F\ar[r]&0}$ be a short exact sequence. Then, for every $E\in R-\Mod$, the sequence $\xymatrix{0\ar[r]&E\otimes_R N\ar[r]^{E\otimes_R f} & E\otimes_R M \ar[r]^{E\otimes_R g} & E\otimes_R F\ar[r]&0}$ is also exact.
\end{lemma}

\begin{proof}


\tcb{type it P 615}
\end{proof}
\begin{proposition}
Let $\xymatrix{0\ar[r]&F'\ar[r]^f & F \ar[r]^g & F''\ar[r]&0}$ be a short exact sequence in $R-\Mod$, and let $F''$ flat. Then, $F$ is flat, over $R$, iff $F'$ is flat, over $R$.
\end{proposition}
\begin{proof}
\tcb{type it P 616}.
\end{proof}
%\input{Topology}
%\backmatter
%\begin{small}
%\input{Bibliography}
%\end{small}
%\printindex
\end{document}